This is a self study question. I am novice at this and have only basic knowledge of solving such problems.

\begin{align} Y &= AK^\alpha H^\beta \\ \dot{K} &= s_KY - \delta K \\ \dot{H} &= s_HY^\psi - \delta H \\ \end{align}

Given, $\alpha + \beta <1$ (DRS) and $\psi \in (0,1]$. Also $A(t) = A(0)e^{gt}$

Objective is to find Balanced Growth path (BGP).

Now I read that from Uzawa's theorem, growth rates of all will be equal in BGP. But the theorem is applicable for only CRS. Since Uzawa is not applicable I am unable to design a good effective capital per labor variable to put $\dot{k}=0$.

By using that growth rates are constant I could only reach till:

\begin{align} (s_k Y/K)/(s_HY^\psi/H) &= constant \\ Y^{1-\psi}H/K &= constant \\ \implies (1-\psi)g_Y + g_H - g_K &= 0 \end{align}

Second equation comes from production function: $g_Y = g + \alpha g_K + \beta g_H$.

Still leaves me with two equations and three variables. Not sure how to go ahead or if this is useful at all.

EDIT: A small clarification: BGP is defined in question as the state when all factors grow at constant rate.

Now based on comments, the second and third equation can be used to get: $g_Y = g_K$ and $g_H = \psi g_Y$. Using these two I can quickly get the BGP path. Now the doubt is whether this approach is right in general? Because I have directly assumed here that BGP exists, i.e., a stable steady path exists. Is this correct??

  • $\begingroup$ Your are missing some optimality condition for the factor inputs, i.e. the relative use of the two capital stocks. $\endgroup$
    – jpfeifer
    Commented Mar 11, 2022 at 7:55
  • $\begingroup$ Do you mean that as a hint or that the question is incomplete? $\endgroup$
    – Dayne
    Commented Mar 11, 2022 at 8:49
  • $\begingroup$ As a hint. Usually, in equilibrium you balance the factor inputs for each unit to have the same net return. That may give you the missing equations/restrictions. Alternatively, try following the logic of Mankiw/Romer/Weil (1992) at eml.berkeley.edu/~dromer/papers/MRW_QJE1992.pdf, Section II.A: Rewrite production as $Y=K^\alpha H^\beta (\tilde AL)^{1-\alpha-\beta}$ with $\tilde A=A^{\frac{1}{1-\alpha-\beta}}$ and $L=1$. At least for $\psi=1$ your model is nested in theirs $\endgroup$
    – jpfeifer
    Commented Mar 11, 2022 at 9:09
  • $\begingroup$ Why do you have $\psi$? The human capital investment is not the same good as consumption? Then what is the level of consumption in your model? $\endgroup$ Commented Mar 11, 2022 at 12:18
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    $\begingroup$ Why write $\left(s_{k} Y / K\right) /\left(s_{H} Y^{\psi} / H\right)$ in a combination? If you do it separately, you will find $g_Y = g_K$. $\endgroup$ Commented Mar 11, 2022 at 12:20

1 Answer 1


As indicated in my comment above, for $\psi=1$, your model is nested in the one of Mankiw/Romer/Weil (1992). You can rewrite production as

$$ Y = A{K^\alpha }{H^\beta } = {K^\alpha }{H^\beta }{\left( {\tilde AL} \right)^{1 - \alpha - \beta }} $$

with $\tilde A= A^{\frac{1}{1-\alpha-\beta}}$ denoting labour-augmenting technological progress and $L=1$. This allows you to divide by $\tilde AL$ to get your system in intensive form:

$$ \begin{align} y &= \left( k \right)^\alpha \left( h \right)^\beta \\ \dot k &= {s_K}y - \left( {\delta + g} \right)k \\ \dot h &= {s_H}y - \left( {\delta + g} \right)h \\ \end{align} $$

In steady state with $\dot k = \dot h =0$, this gives you immediately that the ratios of the capital stocks to output are constant.

But with $\psi<1$, there will be a term depending on technology left in the steady state equation for human capital

$$ 0 = {s_h}y^\psi{\left( {\tilde AL} \right)^{\psi -1}} - \left( {\delta + g} \right)h $$ which is a contradiction. So from what I can see, there is no BGP in this case.

Addendum: To clarify the definition, Acemoglu (2008)'s textbook definition is

By balanced growth, we mean a path of the economy consistent with the Kaldor facts (Kaldor, 1963), that is, a path where, while output per capita increases, the capital-output ratio, the interest rate, and the distribution of income between capital and labor remain roughly constant.

That would not be the case here with respect to human capital.

If you don't care about a balance between factor to output ratios, i.e. a steady state in some appropriately normalized form, then there is at least a steady growth path in this model with $$ {g_Y} = g + \alpha {g_K} + \beta {g_H} = \left( {1 + \alpha } \right)g + \beta {g_H} = \left( {1 + \alpha } \right)g + \beta \psi {g_Y} = \left( {1 + \alpha + \beta \psi } \right)g $$

Income shares with competitive markets where factors are paid their marginal product, in contrast, will be constant due to the Cobb-Douglas assumption. E.g.: $$ \begin{align} {F_K} &= \alpha A{K^{\alpha - 1}}{H^\beta } = \alpha \frac{Y}{K} \\ \frac{{{F_K}K}}{Y} &= \alpha \\ \end{align} $$ But due to non-CRS you need to assume that the rest of output $1-\alpha-\beta$ goes to the fixed factor labor $L=1$

  • 1
    $\begingroup$ What do you mean "a contradiction"? The transform variables of $k$, $h$ could be different when $\psi<1$. $\endgroup$ Commented Mar 11, 2022 at 13:14
  • 2
    $\begingroup$ Following the derivation in the question, we have $g_Y=g_K$, $(1-\psi)g_Y = g_H$, and thus $g_Y = g + \alpha g_Y + \beta (1-\psi)g_Y$. Why isn't this a BGP? $\endgroup$ Commented Mar 11, 2022 at 13:18
  • $\begingroup$ 1. Commonly, a BGP also involves the ratio of capital stocks to output to remain constant. Of course you may have a different definition, but there is no mentioning in the OP. Here, you would have human capital growing slower than physical capital in each period, so the ratio of K/H would go to infinity. 2. If you impose that $\dot h=0$, but something on the right still grows, this is a contradiction. $\endgroup$
    – jpfeifer
    Commented Mar 11, 2022 at 13:31
  • $\begingroup$ Thanks! I will recheck about definition of BGP buy the approach still give me ideas about approaching these problems. $\endgroup$
    – Dayne
    Commented Mar 11, 2022 at 18:02
  • $\begingroup$ @jpfeifer: I have added the clarification regarding the BGP definition. I have also added my approach but I still have a doubt about how to check whether BGP exists. $\endgroup$
    – Dayne
    Commented Mar 13, 2022 at 12:21

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